The resistance of a wire R `Omega` . The wire is stretched to double its length keeping volume constant. Now, the resistance of the wire will become

Options

A. `4ROmega`
B. `2ROmega`
C. `R/2Omega`
D. `R/4Omega`

Explanation

To find the new resistance of a wire after it has been stretched to double its length while keeping the volume constant, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Initial Resistance Formula**:
The resistance \( R \) of a wire is given by the formula:
\[ R = \frac{\rho L}{A} \]
where:
- \( R \) is the resistance,
- \( \rho \) is the resistivity of the material,
- \( L \) is the length of the wire,
- \( A \) is the cross-sectional area of the wire.

2. **Initial Conditions**:
Let the initial resistance of the wire be \( r \). Therefore, we can write:
\[ r = \frac{\rho L}{A} \]

3. **Stretching the Wire**:
When the wire is stretched to double its length, the new length \( L' \) becomes:
\[ L' = 2L \]

4. **Volume Conservation**:
Since the volume of the wire remains constant, we can express the volume before and after stretching:
\[ V = A \cdot L = A' \cdot L' \]
where \( A' \) is the new cross-sectional area. Substituting \( L' \):
\[ A \cdot L = A' \cdot (2L) \]
Simplifying this gives:
\[ A' = \frac{A}{2} \]

5. **New Resistance Calculation**:
Now, we can calculate the new resistance \( R' \) using the new length and area:
\[ R' = \frac{\rho L'}{A'} = \frac{\rho (2L)}{\frac{A}{2}} = \frac{2\rho L}{\frac{A}{2}} = \frac{2\rho L \cdot 2}{A} = \frac{4\rho L}{A} \]

6. **Relate to Initial Resistance**:
We can express this in terms of the initial resistance \( r \):
\[ R' = 4 \cdot \frac{\rho L}{A} = 4r \]

### Conclusion:
The new resistance of the wire after being stretched to double its length while keeping the volume constant is:
\[ R' = 4r \]